In a book I'm reading, the author is optimizing a function with respect to $\sigma^2$. But when finding the maxima and minima of the function, instead of taking the derivative with respect to $\sigma^2$, he does it with respect to $\sigma$! Couldn't that be risky (i.e. extraneous or missing critical points?)
Hey guys, I happen to think that a map f from N to N is order preserving is logically equivalent to saying that f is injective. (I've tried several things and have not find a counter-example, nor can I prove it directly). Any hints?
@anon Sorry, did you mean that as a counter-example? I've also forgot to add that the inequality is < instead of $leq$ in terms of the order preserving definition I'm using
http://math.stackexchange.com/questions/1431731/does-uncountable-sets-always-have-a-countable-dense-subset/1431744#1431744 Can someone tell me why I got a downvote there
R^2 and R^3 are not homeomorphic because R^2 minus a point is not simply connected while R^3 is. but R^2 and R^3 both contain circles, for instance. and so on.
And one more question : If two spaces X and Y are homeomorphic then does that mean $\text{int}(Y)=\emptyset$? I seriously think this is false because because a homeomorphism means a continuous bijection and continuous inverse . So if it is continuous it should take open sets of X to open sets of Y. Now that would imply $\text{int}(Y)\neq \emptyset$@anon
Hello folks, I wish to prove the generalized triangle inequality by induction, but I'm slightly confused with the base case. Do I need to establish the base case for n=1, n=2? since |x1+x2| ≤ |x1|+|x2| has two x terms?
http://chat.stackexchange.com/transcript/message/24028725#24028725 @Balarka When I did this question I started with a space which is not hausdorff and then got the answer. But what I have thought now is that if I find a space which is does not go in accordance with the first countability axiom will not be metrizable . I feel that some quotient group might do the job. In this way we will get better examples than earlier
@DanielFischer Could you have a look at a question for me please in my post. My response to the last comment is as follows:
My idea is as follows: The operator defined by $\Phi_{(z-a)^{-1}}(b):= b(z-a)^{-1}$ is linear continuous operator on $\mathcal{A}$ by continuity of multiplication of Banach algebra $\mathcal{A}$. Let $h(w): = f(w)(w-g(z))^{-1}$. Then
@DanielFischer Also do you know why I can't add what I just wrote as a comment? The comment is not showing the maths it's just showing the latex symbols.
@Moses If you haven't typed a space every now and then, the software inserts zero-width non-joiners every 80 (I believe) characters, and that tends to screw up the MathJax. Use enough spaces in your MathJax. If you have inserted enough spaces, I don't know.
@Moses Okay, you have a mismatch of domains here. There are lots of ways to fix it. You could define $h(w) = f(w)(w-g(z))^{-1}\cdot 1_{\mathcal{A}}$, then $h$ is an $\mathcal{A}$-valued function. More systematic would be to look at continuous linear maps $E \to F$ between any two Banach spaces. The continuity argument for the Riemann integral is oblivious of the specific space(s) involved. Then apply it to $\Phi \colon \mathbb{C} \to \mathcal{A}$ given by $w \mapsto w\cdot (z-a)^{-1}$.
@Bal last lecture we covered some stuff on compact open topology/Arzela-Ascoli, product spaces/Tychonoff, quotient spaces, and classification of surfaces
I don't believe anyone could be stupid. You're just not putting enough effort on understanding, I think.
@morphic well, sounds like you guys are still stuck on boring point-set stuff. how did you do classification of surfaces without knowing fundamental group!?!
If $\tau$ is the standard topology on $[0, 1]$ and $\tau'$ is any other different topology on it, then if $\tau' \subset \tau$, $[0,1]$ cannot be Hausdorff with the topology $\tau'$, and if $\tau \subset \tau'$ then $[0,1]$ cannot be compact with the topology $\tau'$
You start with a solid torus, take another solid torus inside which is knotted inside it but is actually an unknot in the ambient space, and take another solid torus inside and so on. Take intersection.
You get the Whitehead solenoid. Now take complement in R^3.
The resulting thing is called the Whitehead manifold.
@DanielFischer The second systematic idea looks good. The first idea $h(w) = f(w)(w-g(z))^{-1}\cdot 1_{\mathcal{A}}$ involves changing the integrand. Using the second more systematic idea, am I right in observing that all you need to do for any Banach algebra, is to fix any $a \in \mathcal{A}$ then since it is a topological vector space, you have that the mapping $\Phi: \mathbb{C} \to \mathcal{A}$ where $\Phi(w) = w\cdot a$ is continuous linear operator. Then it follows that \begin{align} \Phi\left(\int_{\Gamma_{2}}h(w)dw\right)& =\Phi\left(\lim_{\|\mathcal{P}\|\rightarrow 0}\sum_{j}h(\Gamm…
@DanielFischer Yes I see that now. Thanks. So there is no need to use continuity of multiplication in Banach algebra then. This is better.
@DanielFischer So when you use Riemann-Stieljties integrals and Bochner integrals (I don't really know much about this other than it is similar to Lebesgue integrals but for Banach valued functions) and whatever other types of integrals which you can use to integrate Banach valued functions, these same integrals can be used for the usual scalar valued integrals? Since I guess these are also mappigns into Banach spaces.
@Moses Scalar valued functions are just a special case of vector-valued functions. So the scalar case is subsumed under the vector-valued case. Of course the vector-valued case builds on the constructions for the scalar case, so it's (pedagogically) not a particularly bright idea to start the theory with the vector-valued case.
@DanielFischer Yes I understand. So for all these different types of Banach space valued integrals such as Riemann-Stieljties, Bochner, and some other one that begins with a 'p' that you mentioned previously, they all obviously agree with Riemann integration for integration of real valued functions?
@iwriteonbananas really? it looks like a mess to me, lol
but then i was excited about pathological topological spaces once
i found this one while doing a few exercises from Rolfsen a few months ago
it's a fact that $W \times \Bbb R \cong \Bbb R^4$, even though $W \not \cong \Bbb R^3$, so the cancellation problem in Top, even for manifolds, is false
@Moses It's not the real-valuedness that is crucial. All these integrals agree for "sufficiently nice" functions. The set of integrable functions is different (generally), you should be familiar with that from the Riemann and the Lebesgue integral.
right, i want a non-pathological example of cancellation
this is motivated from the algebro-geometric version of the problem, called the "Zariski cancellation problem". N. Gupta pinned that one down very recently.
@DanielFischer Do you know the name of the integration which is defined on Banach Space $X$ as follows?
Let $I = [a,b] \subset \mathbb{R}$. Let $\mathcal{P} = (t_{k})_{k=0}^{n}$ be a partition of $I$. Let $S(I,X)$ be a normed space of all $X$ valued step functions with the norm $\| \cdot \|_{\infty}$. Then for any $s \in S(I,X)$ we define $$\int_{b}^{a}s(t)dt = \sum\limits_{k=1}^{n}(t_{k}-t_{k-1})x_{k}.$$ Where $\mathcal{P} = (t_{k})^{n}_{k=0}$ is any partition such that $s$ is piecewise constant on $\mathcal{P}$ (such a partition is called admissable for $S$.)
@MikeMiller redo : is every parallelizable manifold orientable in general? is there a way to do this with homology definition?
i recalled that the pushing off orientation by multiplication on Lie group technique was actually something you mentioned in the context of Lie groups being parallelizable.
@Balarka: Here's a sketch of a different proof I just thought of. Let $G$ be a topological group. Show that for any cover $p: G' \to G$ you can give $G'$ a topological group structure such that $p$ is a continuous homomorphism.
@DanielFischer Hi. Could I ask something short. For a direct product of Hilbert spaces from the post, I understand that the direct product of commutative groups $\{ G_{i} \}$ is the full Cartesian product $\prod_{i} G_{i}$, whereas the direct sum $\bigoplus_{i} G_{i}$ is the subgroup of the direct product consisting of all tuples $\{ g_{i} \}$
with $g_{i} = 0$ except for finitely many $i \in I$. Hence for a finite index, the cartesian product and the direct sum are equivalent. If we apply this to Hilbert spaces, are we considering the operation of addition as the commutative group operation of the Hilbert space, thus allowing us to use this result?
you want me to prove that covering space of a topological group is a topological group so that $p$ is a continuous hom? that was an exercise in Munkres.
consider the map $G' \times G' \to G \times G \to G$, first map is $p \times p$ and second being mult. on $G$
@Moses It's more complicated. The direct sum and product as vector spaces of infinitely many (nontrivial) Hilbert spaces are not Hilbert spaces. If you want the resulting space to be a Hilbert space, then products exist only for finitely many (nontrivial) factors, and the construction for the coproduct, the Hilbert sum, is different.
@DanielFischer Just trying to understand the finite direct sum of Hilbert Spaces. Whether I could simply think of the direct product as the set of ordered pairs $H_{1} \oplus H_{2} = \{ (h_{1}, h_{2}): h_{1} \in H_{1}, h_{2} \in H_{2} \}$.
@DanielFischer I meant think of the direct sum as the set...
@Moses Yes, for finitely many (nontrivial) factors, it works like that. You have $H_1 \oplus H_2 \cong H_1 \times H_2$, the inner product is $\langle (x_1,x_2), (y_1,y_2)\rangle_{H_1 \oplus H_2} = \langle x_1,y_1\rangle_{H_1} + \langle x_2,y_2\rangle_{H_2}$, and the obvious generalisation for an arbitrary finite number of Hilbert spaces.
i'm sorry, i haven't thought about it. let me think.
ok, so that cover being topological group part is done.
$\widetilde{G} \to G$ be the orientation double cover of $G$. you give $\widetilde{G}$ a Lie group structure so that the projection becomes a continuous group homomorphism. Now I have to show that $\widetilde{G}$ is the trivial cover. Hmm.
I think one has to use path-tricks here again. That is, I have to try to show that a loop in $G$ lifts to a loop in $\tilde{G}$.
@AndrewThompson: Balarka is trying to prove that a Lie group is orientable. He doesn't know what a tangent space is, so he's trying to use Hatcher's definition. He got stuck with a straightforward proof, so now I suggested an alternate proof by considering the orientation double-cover of $G$.
Hm, I think I can explain tangent spaces reasonably well. (I hope, as I have given several expositions on it on some occasions now.) Want me to give it a try, @BalarkaSen? Might save you some time I think.
@r9m a cute limit I just created $$\lim_{n\to \infty} \left(\sum _{k=1}^n \left(\frac{k (k+1)^k-k^k}{k\text{!!}}\right)\right){}^{\frac{1}{\large n \log (n)}}$$
@robjohn the one above is pretty cute.
@M.S.E I'm pretty involved in finishing my book, and I cannot assign much time to the work on other problems, but later. Yeah, it's an interesting problem and I need to do some more research on some results I got yesterday.
@Balarka I'm blanking here. If $G\to H$ is an onto, many-to-one map of lie groups, and $K\subset G$ are both connected, then is the restriction $K\to H$ also many-to-one of the same degree?
I wouldn't say I think it's true, but I don't know many covering homomorphisms to think of a counterexample
the notion just crossed my mind when trying to show the max torus of SU(2) is a circle (using the fact that's the max torus of SO(3) and the dbl cvr SU(2)->SO(3)).
I suppose we can take $S^1\times S^1\to S^1\times S_1$ which is the double cover in the first coordinate and the identity in the second, then consider $
@Danu Depends. If they clutter the page, then flagging them is good. If they don't clutter and are merely irrelevant, then you can flag or let it be. As long as you don't exaggerate the flagging, we don't care much either way. Dealing with thousands of comment flags per day we would not find thoroughly enjoyable.
@Anthony elements of a direct sum have finitely many nonzero entries, direct product has no such restriction. this distinction is relevant with infinitely many groups or rings or spaces.
Hello, I am unsure whether the main site is best for asking this, so I am asking here. Are there any online math resources/services that provide only text and numbers (no other graphics) and timing similar to what these two services provide:
@Anthony look up product/coproduct. the explanation might not be as powerful as you want, I'm not sure. in any case, consider groups. some are written additively and some multiplicatively, so it makes sense to have both notations. but then with infinite collections there are two different constructions we can talk about, so we might as well reserve one word/notation for one construction and the other for the other.
